Problem 73 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(2 y=\left(\cot ^{-1}\left(\frac{\sqrt{3} \cos x+\sin x}{\cos x-\sqrt{3} \sin x}\right)\right)^{2}, x \in\left(0, \frac{\pi}{2}\right)\) then \(\frac{d y}{d x}\) is equal to : \(\quad\) \(\begin{array}{llll}\text { (a) } \frac{\pi}{6}-x & \text { (b) } x-\frac{\pi}{6} & \text { (c) } \frac{\pi}{3}-x & \text { (d) } 2 x-\frac{\pi}{3}\end{array}\)

Step-by-Step Solution

Answer

Option (b): \( x - \frac{\pi}{6} \).

Step 1: Simplify Cotangent Expression

First, let's simplify the expression inside the inverse cotangent, \( \cot^{-1} \left( \frac{\sqrt{3} \cos x + \sin x}{\cos x - \sqrt{3} \sin x} \right) \). Recognize that this fraction resembles a transformation involving angles. Using trigonometric identities, this simplifies to \( \tan(\pi/6 + x) \). Therefore, \( \cot^{-1} \left( \frac{\sqrt{3} \cos x + \sin x}{\cos x - \sqrt{3} \sin x} \right) = \frac{\pi}{6} + x \).

Step 2: Square the Simplified Expression

According to the problem, \( 2y = \left( \text{cot}^{-1}\left( ... \right) \right)^2 \). From Step 1, substitute \( \text{cot}^{-1}\left( ... \right) \) with \( \frac{\pi}{6} + x \). Thus, \( 2y = \left( \frac{\pi}{6} + x \right)^2 \).

Step 3: Differentiate Both Sides with Respect to x

Differentiate both sides of \( 2y = \left( \frac{\pi}{6} + x \right)^2 \) with respect to \( x \). The right side uses the chain rule: \( \frac{d}{dx} \left( \left( \frac{\pi}{6} + x \right)^2 \right) = 2 \left( \frac{\pi}{6} + x \right) \cdot 1 \). Differentiate \( 2y \) to get \( 2 \frac{dy}{dx} \).

Step 4: Solve for \( \frac{dy}{dx} \)

Setting the derivatives equal gives \( 2 \frac{dy}{dx} = 2 \left( \frac{\pi}{6} + x \right) \). Divide both sides by 2 to isolate \( \frac{dy}{dx} \): \( \frac{dy}{dx} = \frac{\pi}{6} + x \).

Step 5: Match to Given Options

The answer choices need us to find which option correctly represents \( \frac{dy}{dx} = x + \frac{\pi}{6} - \frac{\pi}{6} \). Therefore, we have that \( \frac{dy}{dx} = x - \frac{\pi}{6} \). Looking at the options, this matches option (b).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions help us reverse the process of trigonometric operations to find the original angle. In this exercise, we deal with the inverse cotangent, denoted as \( \cot^{-1} \). This operation allows us to find an angle \( \theta \) when given the value of the cotangent of that angle.
The key is to understand that when you have an expression like \( \cot^{-1} \left( \frac{\sqrt{3} \cos x + \sin x}{\cos x - \sqrt{3} \sin x} \right) \), it can often be simplified using angle transformations or identities.
In this case, by recognizing the angle transformation involved, we simplify it to a more manageable form: \( \tan(\pi/6 + x) \). Knowing how to simplify such expressions using trigonometric identities makes solving inverse trigonometric problems easier.

Trigonometric Identities

Trigonometric identities are mathematical equations that hold true for all angles and provide transformations that simplify complex trigonometric expressions.
In this solution, the identity simplifies the fraction inside the inverse cotangent. Specifically, the expression \( \tan(\pi/6 + x) \) is derived by identifying similarities with known angle sum identities.
The trigonometric identity being used here involves the angles \( \tan(A + B) \) where \( \tan(\pi/6) \) exploits the special angles theorem. Special angles such as \( \pi/6 \) or 30° are frequently used for their simple and known sine, cosine, and tangent values.
This simplification is crucial because it transforms a complex ratio into a simple linear expression, allowing us to further analyze and differentiate it.

Derivative Calculation

Derivatives are foundational tools in calculus used to compute the rate at which a quantity changes. They are used to determine how a function changes as its input changes.
In this exercise, we calculate the derivative \( \frac{dy}{dx} \) from the given equation \( 2y = \left( \frac{\pi}{6} + x \right)^2 \).
To find this derivative, apply the chain rule, a fundamental differentiation rule useful for functions composed of other functions. Differentiating the right hand side, \( \frac{d}{dx} \left( \frac{\pi}{6} + x \right)^2 \), involves multiplying the derivative of the inner function \( \frac{\pi}{6} + x \) by the derivative of the outer function \( 2(\frac{\pi}{6} + x) \), which results in \( 2(\frac{\pi}{6} + x) \).
Once the derivatives are computed, we solve for \( \frac{dy}{dx} \) by isolating it; hence, it leads to the expression \( \frac{dy}{dx} = x - \frac{\pi}{6} \), which matches one of the given answer options.